Diffusion Mass Transfer Chapter 14 Sections 14.1 through 14.7 Lecture 18.

Diffusion Mass Transfer

Chapter 14Sections 14.1 through

14.7

Lecture 18

1. Physical Origins and Rate Equations

2. Mass Transfer in Nonstationary Media

3. Conservation Equation and Diffusion

through Stationary Media

4. Diffusion and Concentrations at Interfaces

5. Diffusion with Homogenous Reactions

6. Transient Diffusion

1. Physical Origins and Rate Equations

1. Driving potential for mass transfer

Concentration Gradient

2. Modes of mass transfer

Convection and Diffusion

General Considerations

Must have a mixture of two or more species for mass transfer to occur.

The species concentration gradient is the driving potential for transfer.

Mass transfer by diffusion is analogous to heat transfer by conduction.

• Physical Origins of Diffusion:

Transfer is due to random molecular motion.

Consider two species A and B at the same T and p, but initially separated by a partition.

– Diffusion in the direction of decreasing concentration dictates net transport of A molecules to the right and B molecules to the left.

– In time, uniform concentrations of A and B are achieved.

• Mass transfer refers to mass in transit due to a species concentration gradient in a mixture.

Definitions

Definitions

iC : Molar concentration of species i. 3kmol/m

i : Mass density (kg/m3) of species i.

i :M Molecular weight (kg/kmol) of species i.

i i iC M

iJ * : Molar flux of species i due to diffusion. 2kmol/s m Transport of i relative to molar average velocity (v*) of mixture.

iN : Absolute molar flux of species i. 2kmol/s m Transport of i relative to a fixed reference frame.

ij : Mass flux of species i due to diffusion. 2kg/s m Transport of i relative to mass-average velocity (v) of mixture.

Transport of i relative to a fixed reference frame.

ix : Mole fraction of species i i ix C C / .

im : Mass fraction of species i i im / .

Absolute mass flux of species i. 2kg/s min:

Mixture Composition

Definitions:

• Mass density of species i: i=Mi*Ci (kg/m3)

• Molecular weight of species i: Mi (kg/kmol)

• Molar concentration of species i: Ci (kmol/m3)

• Mixture mass density: (kg/m3)

• Total number of moles per unit volume of mixture

ii

ii

CC

Mixture Composition

Definitions:

• Mass fraction of species i: mi=i/

• Molar fraction of species i: xi=Ci/C

• For ideal gases:

1 ii

m 1 ii

x

RT

Mp iii

RT

pC i

i ii

pp p

p

C

Cx ii

i

Fick’s Law of Diffusion

For transfer of species A in a binary mixture of A & B

Mass flux of species A (kg/m2s):

Molar flux of species A (mol/m2s):

Binary diffusivity DAB (m2/s)

Ordinary diffusion due to concentration gradient and relative

to coordinates that move with average velocity

AABA mDj

AABA xCDJ *

Fick’s Law of Diffusion

For transfer of species A in a binary mixture of A & B

Mass flux of species A (kg/m2s):

Molar flux of species A (mol/m2s):

If C and are constants, the above equations become:

AABA mDj

AABA xCDJ *

dx

dDDj A

ABAABA

dx

dCDCDJ A

ABAABA *

Mass Diffusivity

p

TDAB

2/3

For ideal gases

Gas-Gas Liquid-Liquid

Gas in Solid

~10-5 m2/s ~10-9 m2/s ~10-10 - 10-13

m2/s

Solid in Solid

~10-19 - 10-34

m2/s

At 298 K, 1 atm

Example 14.1

Consider the diffusion of hydrogen (species A) in air, liquid

water, or iron (species B) at T = 293 K. Calculate the species

flux on both molar and mass bases if the concentration gradient

at a particular location is dCA/dx= 1 kmol/m3●m. Compare the

value of the mass diffusivity to the thermal diffusivity. The mole

fraction of the hydrogen xA, is much less than unity.

Example 14.1

Known: Concentration gradient of hydrogen in air, liquid water, or

iron at 293 K

Find: Molar and mass fluxes of hydrogen and the relative

values of the mass diffusivity and thermal diffusivity

Schematic:

Example 14.1

Assumptions: Steady-state conditions

Properties: Table A.8, hydrogen-air (298 K): DAB= 0.41x10-4 m2/s, hydrogen-

water (298 K): DAB= 0.63x10-8 m2/s, hydrogen-iron (293 K): DAB= 0.26x10-12

m2/s. Table A.4, air (293 K): α= 21.6x10-6 m2/s; Table A.6, water (293 K): k =

0.603 W/m•K, ρ = 998 kg/m3, cp= 4182 J/kg K. Table A.l, iron (300 K): α =

23.1 x 10-6m2/s.

Analysis: Using Eqn. 14.14, we can find that the mass

diffusivity of hydrogen in air at T=293K is )/(10*40.0298

293*)/(10*41.0

298* 2424

298,,

23

23

smsmT

DD ABTAB

Example 14.1

For the case where hydrogen is a dilute species, that is xA<<1, the thermal

properties of the medium can be taken to be those of the host medium

consisting of species B. The thermal diffusivity of water is:

The ratio of the thermal diffusivity to the mass diffusivity is the Lewis number

Le, defined in Equation 6.50.

The molar flux of hydrogen is described by Fick’s law, Equation 14.13.

smKkgJmkg

KmW

c

k

P

/10*144.0/4182*/998

/603.0 263

Example 14.1

dx

dCD

dx

dxCDJ A

ABA

ABA *

Hence, for the hydrogen-air mixture,

25

3

24* 10*41*10*40.0

ms

kmol

mm

kmol

s

m

dx

dCDJ A

ABA

The mass flux of hydrogen in air is found to from the expression:

25

25* 10*810*4*2*

ms

kg

ms

kmol

kmol

kgJMj AAA

Example 14.1

The results for the three different mixtures are summarized in the following table:

Mass Transfer in Nonstationary Media

For mass flux relative to a fixed coordinate system

Mass flux of species A (kg/m2s):

Mass flux of species B (kg/m2s):

Mass flux of mixture (kg/m2s):

AAA vn "

BBB vn "

BBAABA vvnnvn """

BBAA vmvmv Mass-average velocity (m/s):

Absolute Mass Flux

Relative Mass Flux

Mass flux of species A (kg/m2s):

Mass flux of species A (kg/m2s):

AABA mDj

)( vvj AAA

)( vvj AAA vjn AAA "

vmmDvmDn AAABAAABA "

)( """

BAAAABA nnmmDn

Relative Mass Flux

Mass flux of species A (kg/m2s):

Mass flux of species A (kg/m2s):

BBAB mDj

)( vvj BBB

)( """

BABBBAB nnmmDn

0 BA jj

For binary mixture of A & B:

Relative Mass FluxFor binary mixture of A & B:

DAB=DBA

0 BA jj

BA mm

1 BA mm

BBAB mDj

AABA mDj

Absolute Molar Flux

For molar flux relative to a fixed coordinate system

Molar flux of species A (mol/m2s):

Mass flux of species B (kg/m2s):

Mass flux of mixture (kg/m2s):

AAA vCN "

BBB vCN "

BBAABA vCvCNNCvN ""*"

BBAA vxvxv *Molar-average velocity (m/s):

Absolute Molar Flux

Molar flux of species A (kg/m2s):

Mass flux of species A (kg/m2s):

AABA xCDJ *

*)(* vvCJ AAA

**" vCJN AAA

**" CvxxCDvCxCDN AAABAAABA

)( """

BAAAABA NNxxCDN

*)(* vvCJ AAA

Absolute Molar Flux

Mass flux of species B (kg/m2s):

Mass flux of species B (kg/m2s):

BBAB xCDJ *

*)(* vvCJ BBB

)( """

BABBBAB NNxxCDN

0** BA JJ

For binary mixture of A & B:

Example 2

Gaseous H2 is stored at elevated pressure in a rectangular

container having steel walls 10mm thick. The molar

concentration of H2 in the steel at the inner surface is 1 kmol/m3,

while the concentration of H2 in the steel at outer surface is

negligible. The binary diffusion coefficient for H2 in steel is

0.26x10-12 m2/s. what is the molar and mass diffusive flux for H2

through the steel?

Example 2

Known: Molar concentration of H2 at inner and outer surfaces of a

steel wall.

Find: H2 molar and mass flux

Schematic:

Example 2

Assumptions:

Steady-state, 1-D mass transfer conditions

CA<<CB (H2 concentration much less than steel), total

concentration C=CA+CB is uniform

No chemical reaction between H2 and steel

Analysis: )( """

BAAAABA NNxxCDN

Example 2

Analysis: )( """

BAAAABA NNxxCDN

(1). Simplify the molar flux equation

0" BN

1 BA xx

""

11

AAA

ABA

NNx

xxx

Example 2

Analysis:

)( """BAAAABA NNxxCDN

(1). Simplify the molar flux equation

""

" 0

AAA

B

NNx

N

0

Example 2

Analysis:

dx

dCD

dx

dxCDN A

ABA

ABA "

(2). Apply C = constant

tConsCCCC BBA tan

Example 2

Analysis:

tconsdx

dCDN A

ABA tan"

(3). From mass conservation:

)(tan" xftconsN A

AABA dCDdxN "

Example 2

Analysis: (4). Integration from x=0 to x=L, CA=CA,1 to CA,2

2,

1,0

" A

A

C

C AAB

L

A dCDdxN

smkmolxL

CCDN AA

ABA2112,1," /106.2

Example 2

Analysis: (5). Mass flux

smkmolxN A211" /106.2

)/(2*)/(106.2* 211"" kmolkgsmkmolxMNn AAA

)/(102.5 211" smkgxnA

Evaporation in a Column

0"

, dx

dN xA

Evaporation in a Column

Assumptions:

1. Ideal gases;

2. No reaction;

3. xA,0>xA,L, xB,0<xB,L;

4. Gas B insoluble in liquid A

0", xBN

Evaporation in a Column

Stationary or moving medium?

",

", xAA

AABxA Nx

dx

dxCDN

tconsdx

dx

x

CDN A

A

ABxA tan

1"

,

0"

, dx

dN xA

Evaporation in a Column

01"

,

dx

dxdx

xCD

d

dx

dNA

A

AB

xA

Separate variables and integrate

21)1ln( CxCxA

Evaporation in a Column

Apply B.C.’s to solve C1 & C2:

21)1ln( CxCxA

LAA

AA

xLx

xx

,

0,

)(

)0(

Lx

A

LA

A

A

x

x

x

x /

0,

,

0,

)1

1(

1

1

Lx

B

LB

B

B

x

x

x

x /

0,

,

0,

)(

)1

1ln(

0,

,",

A

LAABxA x

x

L

CDN

Example 3

 An 8-cm-internal-diameter, 30-cm-high pitcher half

Filled with water is left in a dry room at 15°C and

87 kPa with its top open. If the water is maintained

at 15°C at all times also, determine how

long it will take for the water to

evaporate completely.  

Example 3

Example 3

Example 3

Lecture 18